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1:01 PM
what in the world are we talking about?
 
> EDIT: Ok, so the entire system is a huge mess.
from
14
Q: Why are journals used in modern academic research?

Souradeep NandaI am new to research and I have yet to internalize the concept of journals and their utility in archiving scientific literature. Almost all the papers I have read recently are from this website called arxiv. Arxiv calls itself to be a preprint archive. Anyone can upload a PDF file to the reposit...

nicely summed up.
 
there's also this paper on the topic
I think those are the only two?
I should look up INSPIRE HEP
Not referenced there
dang it
well technically it's a math paper
the other paper does specify that the path of a sting is specifically arbitrary
that paper never even mentions why stings are to be added
"Wolfgang Smith (born 1930) is a mathematician, physicist, philosopher of science, metaphysician, Roman Catholic and member of the Traditionalist School. He has written extensively in the field of differential geometry, as a critic of scientism and as a proponent of a new interpretation of quantum mechanics that draws heavily from medieval ontology and realism."
also apparently the author is insane
 
@EmilioPisanty arXiv of course is a really big deal in hep-th and actual journals are more just to satisfy pointy-haired bureaucrats :P
2
interestingly, I was talking to a computer science guy a few weeks ago, since I recently wrote a more CS-oriented paper
he told me that nobody in that field cares about the arXiv
 
@BenNiehoff yeah, only there's more to science than hep-th, much as it pains hep-th'ers to admit it
=P
 
1:13 PM
and if you peruse the CS sections of arXiv, it shows...there is not much there, and not of great quality
 
yup. Same with very many subfields.
Show me a quantum chemist that cares about the arXiv and my jaw will drop to the floor
 
It's unfortunate
 
I'm curious whether the importance drops even for closely related fields like hep-ph or hep-ex...
 
Arxiv is pretty useful
 
what's hep-ph?
 
1:14 PM
well, most fields of science deal with things that are actually observable ;)
 
Trying to look at papers when you're not in a lab for a field that's not on arxiv is pretty harsh
Like looking for a history paper?
Forget about it
 
so they have the opportunity to publish in places like Nature or Science, and win Nobel prizes
 
@Slereah you can always follow the lure of the siren's call
where of course the siren has initials S-H.cc
 
the ph in hep-ph means phenomenology
 
hush
that is a secret
 
1:16 PM
@Slereah if it's on the NYT it's not a secret
what does seem to be a secret is the scihub Telegram bot
mighty useful, that
 
yeah but if you bring it up too much you might get censored :p
 
@BenNiehoff wtf is hep-phenomenology
@Slereah what, here on this chatroom?
that'd be interesting to see
I have seen institutions get pretty touchy when their members start using scihub too much
leading to network blocks and so on
which is where the Telegram bot comes in
end-to-end encrypted, so it's a transaction strictly between you and the raven
 
@EmilioPisanty i read through (okay, skimmed) the articles linked...i knew there were problems, but i didn't realize it was that nuts.
 
@heather it is completely insane, yes.
in that spirit, though, may I suggest you consider upvoting this one? It's one away from going live
 
@BenNiehoff some physics fields don't even care about TeX...
 
1:25 PM
Man I can't find any paper referencing that paper
it's one of those GHOST PAPERS
 
@EmilioPisanty 'tis upvoted
 
@heather much 'ppreciated
@0celo7 that's a good definition for "no longer physics"
 
@EmilioPisanty oh please
@ACuriousMind I have to leave today, my poor cat will be all alone again
 
@0celo7 Yes, I know biophysics people who submit papers in Word format
 
@0celo7 above statement has a 0% tongue-in-cheek content
 
1:27 PM
@0celo7 :(
 
@0celo7 how to prove that the causal diamond in minkowski space is bounded
I guess mb switch to null coordinates and then show it's in a cube?
 
"the" causal diamond?
 
@BenNiehoff I want to say that there are some journals that prefer Word, but I don't know for certain.
 
Well, all causal diamonds
 
or "a" causal diamond?
 
1:28 PM
@Slereah bounded in which sense
 
In the Heine-borel sense
 
I wonder whether my diamond is causal...
 
If you're trying to use Heine-Borel, that won't work
It doesn't work on every manifold
"Bounded" depends on the metric
 
Well it's Minkowski space
 
i'm reading scott aaronson's paper, and it's rather fabulous =)
 
1:29 PM
The most $\Bbb R^n$ of all manifolds
 
@Slereah you can give an explicit expression for the causal diamond then
It won't be nice, but is certainly possible
 
Yeah I guess I can probably show that it's bounded by some cube
 
But why are you doing this? Just show it has a Cauchy surface.
 
Would it be easier?
 
@ACuriousMind 💍
@Slereah probably
 
1:30 PM
@0celo7 ...what is that?
 
Wald or HE has some characterization of Cauchy surfaces using null goedeiscs
Null goedeiscs in Minkowski space are easy
 
Oh i guess that would be simple yeah
 
Oh, it's a...ring emoji? It renders weirdly in my browser
 
@ACuriousMind a diamond ring emoji
 
1:32 PM
Huh
@ACuriousMind Did your FA course cover TVS in full generality?
 
@0celo7 No
We rather quickly went to the setting of Banach spaces and Hilbert spaces
 
Those are easy. It's surprisingly hard to prove simple things in general TVS
 
I'm curious, how frequently on StackExchange do you get people who seemed to have learned a whole lot of fancy words, but don't seem to understand basic concepts, and then they get super argumentative...
 
@BenNiehoff that's basically everyone on PSE
 
@BenNiehoff Too frequently :P
 
1:34 PM
haha
oh, and throw in a language barrier for extra fun
so that all they can really do is repeat themselves
with the fancy words
 
I won't name anyone but his name rhymes with John Muffield
 
@EmilioPisanty
1
Q: Transition rate of two level system subjected to noise

DanielSank(this question is simpler than its length implies. I did this on purpose to provide a nice complete development for future readers) The setup Suppose we have a two-level quantum system with Hamiltonian $$H/\hbar = -\frac{\Omega}{2} \left[ I(t)\sigma_x + Q(t) \sigma_y \right] \, .$$ Now suppos...

I was surprised nobody ever answered that one.
 
@BenNiehoff For real, physicists do have a tendency to use fancy words. My QM prof keeps calling the Dirac delta a distribution, but then he uses it like a function and on test functions that are not even test functions
 
the best test function is Dirac delta, of course!
 
Either use the incorrect terminology to fit your usage of the concept or use everything correctly
 
1:38 PM
maybe complex conjugated
 
Fucking Taylor expanded Dirac delta
I'm still mad about that
 
@DanielSank no idea about that one
me and noise don't get on very well
 
wait, what? how do you Taylor expand Dirac delta?
 
or are you sneakily soliciting for bounties?
 
2 days ago, by AccidentalFourierTransform
@0celo7 have fun
Start there
 
1:39 PM
@BenNiehoff please let's not go into that again
 
@BenNiehoff Write the delta as a limit of something else, Taylor expand, and then take the limit?
@EmilioPisanty I'd hardly call it sneaky.
 
@DanielSank hello =)
 
How do you project an operator onto a subspace?
 
I didn't get an answer for my Majoranas even after a 500 point bounty :(
 
@DanielSank What does "constant spectral density" even mean?
 
1:41 PM
Suppose I have operator $X$ and I want to project onto a space spanned by $|a \rangle$ and $|b\rangle$.
 
and how is it different from a signal with infinite power?
 
@ACuriousMind I know people who know Majoranas...
@EmilioPisanty It means the spectral density is constant with frequency.
 
@DanielSank Take the projector onto the subspace and multiply the operator from both sides with it.
 
Whoa whoa whoa
 
@ACuriousMind Oh, did your bounty expire? I was thinking of looking into that one, but I was busy last week and I'm not sure if I'd come up with a more definitive answer than Polchinski and Figueroa-O'Farrill anyway :P
 
1:41 PM
@EmilioPisanty It's not. Normally if someone says "constant spectral density" they mean that it's constant over some range of itnerest.
 
Is @DanielSank needing functional analysis advice?
 
@BenNiehoff It expired a month ago
 
$$\left( |a \rangle \langle a | + |b \rangle \langle b | \right) X \left( |a \rangle \langle a | + |b \rangle \langle b | \right) $$
 
lol
 
^ That?
@0celo7 No.
 
1:42 PM
And I'd be happy with Polchinski and O'Farrill if they didn't contradict each other!
@DanielSank Yep.
 
Never mind then. I need to get packing
 
@heather Hi.
 
@DanielSank If they can answer this, I'd be very happy
 
Figueroa-O'Farrill is honestly probably the weirdest hyphenated name I've seen :P
 
@ACuriousMind There's already an answer with lots of votes.
 
1:44 PM
Heh, yes, that is true :P
Unfortunately it's the wrong kind of answer and the wrong kind of votes for me to be happy
 
lol, I love that answer
 
@ACuriousMind d'you reckon a second bounty would help?
 
"They don't exist, because they're math, and we don't do math here!"
 
@EmilioPisanty I fear it would not
 
Oh man
 
1:45 PM
@EmilioPisanty This is a case for targeted advertising.
 
@ACuriousMind yeah, I'm afraid of that
 
@ACuriousMind This is more mathy than I expected. Let me think if I know anyone who could answer this.
 
I really considered adding one immediately after yours expired
 
> Hence I await my 500 point bounty.
that comment he made is just so...
 
but I was already steeped in enough animosity from the answerer
 
1:46 PM
i've marveled both times i've read it
 
Best impulse buy ever:
 
wait, wait, wait - look at the vote difference. How the heck did he get 5 upvotes on that answer?!
 
I really need to get around to reading that
 
@ACuriousMind I had a friend in grad school who would maybe be able to help (Polchinski's student).
 
@DanielSank You expected something less mathy from me? :D Not sure how to take that
 
1:47 PM
I'm not sure how it would go at this point for me to send him a SE question.
 
@heather One of the continuing mysteries of this site
 
He works in a biophysics lab now.
Like, in the lab.
 
2nd volume has chapters on diff geom, algebraic geometry
1st volume has modular forms, $E_8 \times E_8$ insanity
 
It also means that he made a net +2 rep, even from an answer as spectacularly off-topic as that one
 
Way better than the paperback too!
 
1:48 PM
I do remember that in some dimensions one has two different choices of charge conjugation matrix, and maybe they are not always equivalent
 
@DanielSank Haha, good for him (assuming he likes it)
 
@ACuriousMind I think he does.
 
so that could at least explain the discrepancy between Joe and José
 
Not sure how his Majorana-fu is nowadays.
 
but I dunno for sure
 
1:49 PM
The people I know now are mostly condensed matter theorists who propose experiments to actually detect the particles.
Turns out superconducting loops with bits of high spin-orbit coupling semiconductors might be a way to see them.
 
Hm, while they have no reason not to know I fear CM theorists have little reason to care about general Lorentz representations
 
@bolbteppa you got your prerequisites on hand?
 
@ACuriousMind Correct.
Oh well.
 
@heather I have him one for a good attempt
 
haha
Honestly, I found the closest thing in a google translate of a Lubos course where he references GSW
For the $E_8$ stuff
 
1:52 PM
@BenNiehoff do you know the Fourier expansion of the delta?
 
Which one
 
The usual one
 
@Kaumudi.H Did you appear for the Resonance tests ?
 
$$ \int \frac{d \omega}{2\pi} \exp(i \omega t) = \delta(t)$$
 
@0celo7 Yes, of course
but of course the integral doesn't really converge
 
1:55 PM
@BenNiehoff Regularize it.
 
and neither delta nor its Fourier transform belong to Hilbert space, etc.
@DanielSank Well sure. Prove that the answer is unique no matter how I regularize it :)
 
You know $\lim_{x\to0}\frac{sin(x)}{x} = 1$, from this you can get your delta function if you push it
 
@bolbteppa Oh, I also have an open question about $E_8$ stuff
 
@BenNiehoff Indeed. Normally the choice of regularizer comes from boundary conditions.
 
@BenNiehoff if you're a physicist it does
Now imagine inserting id/dx for x in the integral
Now you have delta(id/dx)
 
1:57 PM
(While I'm at the shameless self-advertising :P)
 
It's a wonderful operator if you pretend it makes sense
 
@0celo7 Most physicists I know realize that there are certain technicalities involved...somehow they manage to end up with the right answers, though
 
@BenNiehoff The only tricky bit is which way to regularize.
 
@BenNiehoff I want to ask a question on PSE about when that fails
 
@ACuriousMind I have a question about path integration.
 
1:59 PM
I'm still only on the Atiyah Comm Alg level :p
 
@bolbteppa there's plenty of functions $f$ such that $\lim_{x\to0}f(x) = 1$ and which don't yield delta functions the way $\mathrm{sinc}$ does.
 
I am working with Wiener path integrals, which have the basic case $$\int_{\mathcal{C}(x,t;0,0)} d_W[x(\tau)] = \frac{1}{\sqrt{4 \pi D t}} \exp \left( - \frac{x^2}{4 \pi D t} \right) $$
 
Sure
 
In other words, we're working with diffusion.
With me so far?
 
2:01 PM
@0celo7 I think it fails when the operators you're talking about are somehow badly-behaved, but I don't know the details
 
Although I have to say I don't understand the relation to diffusion/Brownian motion very well
 
I think it is similar to when Dirac bra-ket notation fails
 
Here $\mathcal{C}(x_f,t_f;x_i,t_i)$ is the set of paths starting at $(x_i, t_i)$ and ending at $(x_f, t_f)$.
 
it's basically all Dirac's fault!
 
@ACuriousMind Well, the relation is that if I put a particle at $(0,0)$ and let it diffuse, it's probability distribution at time $t$ is $$\frac{1}{\sqrt{4 \pi D t}} \exp \left( - \frac{x^2}{4 D t} \right) \, . $$
 
2:03 PM
Physicists: Pretending the Hermitian conjugate is the adjoint since 1932
 
@ACuriousMind ok?
(I messed up a $\pi$ in the first equation)
 
@DanielSank Yeah, okay. I feel there's something deeper there, but okay
 
@ACuriousMind There really isn't.
The Wiener path integral is constructed to give the right answer for diffusion. In other words, you define the measure to make sure you get diffusion.
Ok so now suppose I want to actually evaluate some path integrals.
One way to do this is with a finite approximation.
In between $t_i$ and $t_f$, I introduce $N$ points spaced by $\epsilon$.
 
Mhm...then what?
 
typing...
The probability that a path goes through a set of "gates" at those times is $$\int dx_1 \cdots dx_N \frac{1}{\sqrt{4 \pi D \epsilon}^N} \exp \left( - \frac{1}{4 D \epsilon} \sum_{i=1}^N \left( x_i - x_{i-1} \right)^2 \right) \, . $$
 
2:10 PM
@TheDarkSide If you really care about it - some of us have written to him directly expressing hope that he will come back, though to my knowledge he hasn't replied to any such contacts. Still, if done respectfully and once, it's something to consider.
 
Ok?
 
@DanielSank Yes
 
Great. Now, suppose we want to know the transition probability from $(0,0)$ to $(0,0)$.
After a total time $\tau$ or something.
In that case, we note that the sum is a quadratic form and write it as $\sum_{i,j} x_i x_j A_{ij}$.
Then we just have a Gaussian integral and we know the result is $$\frac{\sqrt{\pi}^N}{\sqrt{\text{det} A }} \, .$$
 
I found the best linear algebra book of them all, Halmos, looking over it, definitely the best!
 
@DanielSank Yep
 
2:14 PM
We can do this because $x_0 = x_{N+1} = 0$, so that original sum involves only the integration variables etc. etc.
We can even evaluate $\text{det}A$. It turns out to be $N+1$.
How do we compute the transition probability from $(0,0)$ to $(x,t)$?
(Ooops, there's a mistake above. It should have been transition probability from $(0,0)$ to $(0,t)$ in the first problem)
If the final position is not 0, it doesn't vanish from the sum in the exponent.
 
Look at how he introduces inner products
 
@bolbteppa "We assert" is never enticing, IMHO.
 
halmos wrote a linear algebra book? didn't realize that (i have his set theory book)
 
Can't find him say that
 
How do we deal with having a non-integrated variable in there?
 
2:18 PM
@DanielSank Well, but it's then a Gaussian integral with a linear term, no? So complete the square.
 
He also wrote a problem book on linear algebra to go with it
 
@ACuriousMind Ah! Right... I've heard of this...
 
From the viewpoint of tbe integral, $x_{N+1}$ is just a number
 
Yes but it's coupled to the variables I'm integrating.
So I have to write some kind of $J \cdot x$ term?
 
Wait a moment
 
2:19 PM
k
 
The square is already completed. Can you not just do the transformation $x_N\to x_N + x_{N+1}$ and get the original integral?
 
He even makes a great point about the JNF coming from isolating the most invertible and most zero-ness of any map, though the image and null space not being disjoint means you have to modify your notion of zero-ness to nilpotency
 
Well, going back a step...
 
I think that works. Go back to where you have the sum $\sum_i (x_i - x_{i-1})^2$ and do the transformation $x_N\to x_N + x_{N+1}$
Then the $x_{N+1}$ vanishes from the integrand and you're fine
 
The sum looks like this: $$x_0^2 - x_1x_0 + x_1^2 + x_2^2 + x_1^2 - 2 x_2 x_1 + \cdots + x_{N+1}^2 - 2 x_{N+1}x_N \, . $$
 
2:23 PM
 
You're saying to define $y \equiv x_N - x_{N+1}$?
 
@DanielSank Yeah
Oh, no, wait
 
(That $\rightarrow$ notation always confuses me)
 
That gives you the same problem in the term $(x_N -x_{N-1})^2$
Nevermind
 
@ACuriousMind Yep.
 
2:24 PM
But, anyway, you've got a $x_N^2$ in there, right?
 
@bolbteppa man, I miss abstruse goose
 
Well, it winds up being $2 x_N^2$.
 
So gather all terms with $x_N$ in it, you get something of the form $ax_N^2 + b x_N$
 
Uhhh
You have $$2x_N^2 - 2 x_N x_{N-1} - 2 x_N x_{N+1} \, .$$ among other terms.
$x_{N+1}$ is the one that's not a variable.
Here, let me write out the last few terms of that sum...
 
stopped updating in like 2014 =(
 
2:27 PM
@DanielSank Ah, yes, I see
 
Damn!
 
$$-2 x_N x_{N-1} + 2 x_N^2 + x_{N+1}^2 - 2 x_N x_{N+1}\, . $$
In the problem where we end at $x=0$, $x_{N+1}=0$ so we drop the troublesome terms.
 
is there a tag for basis-states, of i.e., a quantum system?
 
@bolbteppa yeah, it's a shame
 
there doesn't seem to be, but maybe i'm searching the wrong things.
 
2:30 PM
@heather You define your own.
For example, by convention, if you're working with a harmonic oscillator, then $|n \rangle$ refer to the energy basis states.
 
i understand, i meant a tag for a question about them (on PSE).
there seems to be a tag for
 
@heather Every state can conceivably be chosen as part of a basis, I'm not sure that's a useful tag
@DanielSank This problem is surprisingly tricky, or I'm just rusty with integrals
 
@ACuriousMind I like to think I don't ask for help on easy stuff...
 
The last few terms don't really help - since each $x_i$ is coupled to both $x_{i-1}$ and $x_{i+1}$, every change you make there will propagate through the entire sum, so to speak
 
@ACuriousMind Right. Is there a way to handle this? I figured this were standard operating procedure for you mathy folks.
(How do I get magic chat edit powers? I need that.)
 
2:37 PM
The problem with the $x_{N+1}$ appearing is that this has no real nice interpretation anymore. In the $x_{N+1} = 0$ case, you can just write the integral as $\int_{\mathbb{R}^N} \exp(x^T A x)\mathrm{d}x$.
 
Right, so what if we look at the $x_{N+1}^2$ term as "separate" and look at the $-2x_n x_{N+1}$ term as a source term?
I'm just learning about these source term thingies again. I presumably learned them in a course many moons ago...
 
Once it is non-zero, it becomes $\int_{R^N}\exp(x'^T A x')\mathrm{d}x$ where $x'$ is now $x + x_{N+1}$.
 
Wait a sec... the $x_{N+1}^2$ term isn't coupled to anything, right? Can't we separate it as a prefactor?
 
$$ \int \exp(x A x + x_{N+1} A x + x^T A x_{N+1} + x_{N+1} A x_{N+1})$$
 
So, I have this guy I was trying to help on PhysicsForums...he has started going on about jet bundles and Gribov ambiguities, but doesn't seem to understand that the Levi-Cività connection and gauge connections are basically the same type of object
 
2:41 PM
@ACuriousMind Wat?
Lemme ask this, is there a formula for $$\int dx_1 \cdots dx_N \exp \left( - A_{ij}x_i x_j + J_i x_i \right) \, ? $$
 
@DanielSank Yes. You complete the square, you get the result without source times $\exp(\frac{1}{2} J A^{-1} J)$
 
Oh! Ok then I think we can solve this bad boy.
 
Your "source" here would be $J_i = 2 A_{i,N+1} x_{N+1}$
 
Er... well... If I can figure out $J A^{-1} J$.
 
And you have an additional factor $\exp(x_{N+1}^2)$, but that's not a problem.
So, yeah, you have to compute the inverse of your $A$.
@DanielSank I'm afraid you'll have to become a mod ;P
 
2:49 PM
@ACuriousMind I'm not sure what you mean here but I think I get the idea.
@ACuriousMind Ugh. In this case, $A$ is the discrete version of the second derivative, i.e. $\nabla^2$. Surely someone knows the inverse of it?
$$A_{N=3} = \left[ \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} \right]$$
i.e. 2 on the diagonal and -1 on the first off-diagonals.
 
@DanielSank If you compare your formula with the once you just asked me about, you find that the $J_i$ are $2 A_{i,N+1}x_{N+1}$ for $i = 1\dots N$, which, come to think of it, is just $-2 \delta_{i,N} x_{N+1}$.
 
@ACuriousMind Right. I was wondering what was up with the $A$ stuff when it's just a delta.
 
@DanielSank Surely, but I'm afraid I only know it's the propagator in the continuous case, no idea about the discrete
 
@ACuriousMind Well, I intend to take a continuum limit at the end...
Enlighten me.
All this discrete stuff is just a crutch because I'm still learning.
 
This place has been there for years and no lawsuit has brought it down.
 
2:54 PM
@DanielSank Well, it's almost a tautology - the propagator is the Green's function for $\nabla^2$, and taking the integral with the Green's function is the inverse operator to the operator it's the Green's function for by definition
 
@ACuriousMind Right.
So is $JA^{-1}J$ the transition probability (amplitude in QM) from the source back to the source?
 
I'm not sure I understand what you mean by that. It's the additional factor you get when you complete the square in the Gaußian, I don't have a good interpretation for it
 
Suppose we have something that evolves according to $\nabla^2$.
We do this path integral business and wind up with $J (\nabla^2)^{-1} J$.
 
I think I got what you want to say and yes, it is
 
I suspect there's a simple way to understand this.
@ACuriousMind Ok.
I think that $J (\nabla^2)^{-1} J$ is the overlap between the "state" represented by the source and the propagated version of that initial state.
Does that sound right?
 
2:59 PM
Yes, exactly
 

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